Telangana SCERT Solution Class X (10) Maths Chapter 8 Similar Triangles Exercise 8.1
(1) In ∆PQR, ST is a line such that PS/SQ = PT/TR and also ∠TSP = ∠PRQ. Prove that ∆ PQR is an isosceles triangle
(2) In the given, LM II CB and LN II CD, Prove that AM/AB = AN/AD.
Given : LM II CD and LN II CD
To Prove : AM/AB = AN/AD
proof :- In ∆ABC
LM II CB
=> AL/LC = AM/MB (by basic proportionally Theorem)
In ∆ADC,
LN II CD
=> AL/LC = AN/ND (by BPT)
From (1) and (2)
AM/MB = AN/ND
MB/AM = ND/AN
add 1 on both sides,
MB/AM+1 = ND/AN+1
MB+AM/AM = ND+AN/AN
AB/AM = AD/AN
=> AM/AB = AN/AD
Hence Proved
(Q3) In the given figure, DE II AC and DF II AE, Prove that, BF/FE = BE/EC
Given : DE II AE and DF II AE
To Prove: BF/FE = BE/EC
Proof : In ∆ABC
DE II AC
by basic proportionality Theorom
BD/DA = BE/EC ………….. (1)
From Equation (1) & (2)
BE/EC = BF/FE
=> BF/FE = BE/EC
Hence Proved
(Q4) Prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side (using basic proportionality theorem)
(Q5) Prove that a line joining the midpoints of any two sides of a triangle is parallel to the third side. (Using converse of basic proportionality theorem)
DE = BC
Hence proved.
(Q6) In the given figure, DE || OQ and DF || OR. Show that EF || QR
by Converse of B.P.T
EF II QR
(Q7) In the adjacent figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC||PR. Show that BC || QR.
From equn (1) & (2)
OB/BQ = OC/CR
By Converse of B.P.T
BC II QR
Hence Proved
(Q8) ABCD is a trapezium in which AB||DC and its diagonals intersect each other at point ‘O’. Show that AO/CO =BO/DO
(Q9) Draw a line segment of length 7.2 cm and divide it in the ratio 5 : 3. Measure the two parts and verify the results.
Here is your solution of Telangana SCERT Class 10 Math Chapter 8 Similar Triangles Exercise 8.1
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